LeetCode笔记:137. Single Number II
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问题:
Given an array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
大意:
给出整数数组,每个元素都出现三次,只有一个元素只出现一次,找到它。
注意:
你的算法应该是线性时间复杂度。能不用额外的空间来做吗?
思路:
用哈希表也可以做到线性时间复杂度,但是用了额外空间。
这里我们参考这个思路来做:传送门:算法题总结之找到数组中出现次数唯一不同的数字
代码(Java):
public class Solution { public int singleNumber(int[] nums) { int x1 = 0; int x2 = 0; int mask = 0; for (int i = 0; i < nums.length; i++) { int num = nums[i]; x2 ^= x1 & num; x1 ^= num; mask = ~(x1 & x2); x2 = x2 & mask; x1 = x1 & mask; } return x1; }}
合集:https://github.com/Cloudox/LeetCode-Record
版权所有:http://blog.csdn.net/cloudox_
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